Greg has an array a?=?a1,?a2,?…,?an and m operations. Each operation looks as: li, ri, di, (1?≤?li?≤?ri?≤?n). To apply operation i to the array means to increase all array elements with numbers li,?li?+?1,?…,?ri by value di.
Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1?≤?xi?≤?yi?≤?m). That means that one should apply operations with numbers xi,?xi?+?1,?…,?yi to the array.
Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.
Input
The first line contains integers n, m, k (1?≤?n,?m,?k?≤?105). The second line contains n integers: a1,?a2,?…,?an (0?≤?ai?≤?105) — the initial array.
Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1?≤?li?≤?ri?≤?n), (0?≤?di?≤?105).
Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1?≤?xi?≤?yi?≤?m).
The numbers in the lines are separated by single spaces.
Output
On a single line print n integers a1,?a2,?…,?an — the array after executing all the queries. Separate the printed numbers by spaces.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.
Sample test(s)
Input
3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3
Output
9 18 17
Input
1 1 1
1
1 1 1
1 1
Output
2
Input
4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3
Output
5 18 31 20
用一个小技巧就行了,线性复杂度
/*************************************************************************
> File Name: CF-179-C.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2015年04月26日 星期日 21时52分26秒
************************************************************************/
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
static const int N = 100100;
LL arr[N];
LL cnt[N];
LL cnt2[N];
struct node {
int l, r;
LL d;
}ope[N];
int main() {
int n, m, k, l, r;
LL d;
while (~scanf("%d%d%d", &n, &m, &k)) {
memset(cnt, 0, sizeof(cnt));
memset(cnt2, 0, sizeof(cnt2));
for (int i = 1; i <= n; ++i) {
scanf("%I64d", &arr[i]);
}
for (int i = 1; i <= m; ++i) {
scanf("%d%d%I64d", &ope[i].l, &ope[i].r, &ope[i].d);
}
while (k--) {
scanf("%d%d", &l, &r);
++cnt[l];
--cnt[r + 1];
}
for (int i = 1; i <= m; ++i) {
cnt[i] += cnt[i - 1];
}
for (int i = 1; i <= m; ++i) {
l = ope[i].l;
r = ope[i].r;
d = ope[i].d;
cnt2[l] += d * cnt[i];
cnt2[r + 1] -= d * cnt[i];
}
for (int i = 1; i <= n; ++i) {
cnt2[i] += cnt2[i - 1];
}
printf("%I64d", cnt2[1] + arr[1]);
for (int i = 2; i <= n; ++i) {
printf(" %I64d", cnt2[i] + arr[i]);
}
printf("\n");
}
return 0;
}